Moving Maximum in last 5 minutes in R - r

Question

I was wondering how to implement a moving maximum and minimum for a price in the last 5 minutes in O(n) time in R. My data consists of two columns: one with the time of day in seconds and the other with price. Right now, I take the current time, subtract 5 minutes, subset for the last 5 minutes, and then search for min and max at each index, so the operation is O(n^2). Is there any way to do this in O(n)?
Sample data:
time
[34200.19, 34200.23, 34201.45, ..., 35800, 35800.2, 35800.5]
price
[100, 103, 102, ..., 95, 97, 99]

Answer 1

The following compares a direct approach with a slightly more efficient varient, but it looks to scale as about n^1.6 on the values I've tried it with (10,000 - 100,000) - partly depends if incresing n is assumed to be more points in the same time period, or extending over a longer period.
#Create some data
n <- 10000
d <- data.frame(t=as.POSIXct(24*3600*runif(n), origin = "2014-01-01"),x=runif(n))
d <- d[order(d$t),]
d$inmax2 <-d$inmax <- rep(FALSE,n)
d$inmax2[1] <-d$inmax[1] <- TRUE
if (max(diff(d$t)) > 300) warning("There are gaps of more than 300 secs")
#Method 1, assume that you've done something like this
t1 <- system.time({
for (i in 2:n) d$inmax[i] <- !any((difftime(d$t[i], d$t[1:(i-1)] ,units="secs") < 300) & (d$x[i] < d$x[1:(i-1)] ))
})
#Method 2
t2 <- system.time({
cand <- 1
next_cand <- 2
while (next_cand <= n)
{
cand <- cand[difftime(d$t[next_cand],d$t[cand],units="secs")<300]
cand <- c(cand[d$x[cand] > d$x[next_cand]],next_cand)
if(length(cand)==1) d$inmax2[cand] <- TRUE
next_cand <- next_cand + 1
}
})
rbind(method1=t1,method2=t2)
# user.self sys.self elapsed user.child sys.child
# method1 14.98 0.03 15.04 NA NA
# method2 2.59 0.05 2.63 NA NA
all(d[[3]]==d[[4]])
# TRUE
The approach is to run through keeping all possible candidates in the past 5 minutes that are not less than the present one. If there are no such candidates the current must be the maximum. I assume that you can generalise to minimum.
Possibly doesn't work if you want to know maximum in last 5 minutes between datapoints rather than at datapoints though - not sure if you require that

Answer 2

Sort the dataframe by time first. Then maintain a max heap of the price, removing the lost price entries after every shift. Since rebalancing a heap is O(log n), this will be O(n log n). For implementing a max heap, consult any algorithms textbook (although I may edit this post later with one).